Printing an array of characters

Question

I was curious about putting strings in a pointer of a character, so I tried this out:

int main() {
char *str = "Solo Learn";
for (int i = 0; i < 100; i++){
cout << i << ": " << str[i] << endl;
}
cout << "Program completed.";
return 0;
}

I arbitrarily put i<100 because I was curious, and then the result ended up as if the string were
"Solo Learn : Program completed. C POSIX basic_string::erase 7C α6C ╖6C ó6C 7C Q0C 00C 0C ≥/C "
Then my "Program completed." line showed up.

Why does it pick up all other strings in the program, and where do the other characters come from?

Answer 1

C++ does not have an automated control system for beeing outside of your String...See following example

This is the memory around and with your String.

Def. | memory space | ; x is a filled memory space with something we dont know

... | x | x | 'S' | 'o' | 'l' | 'o' | ' ' | 'L' | 'e' | 'a' | 'r' | 'n' | x | x | ...

By printing from i to 100 you are doing following:

cout << str[0]; // prints S
cout << str[1]; // prints o
...
cout << str[9]; // prints n

cout << str[10]; // prints now the content of the x stored behind your String. See here:

... | x | x | 'S' | 'o' | 'l' | 'o' | ' ' | 'L' | 'e' | 'a' | 'r' | 'n' | x | x | ...
^
str[10]
and so it is going on... That is the crazy stuff behind your String in the output result you got